Description#
You are given a positive integer n representing the number of nodes in a connected undirected graph containing exactly one cycle. The nodes are numbered from 0 to n - 1 (inclusive).
You are also given a 2D integer array edges, where edges[i] = [node1i, node2i] denotes that there is a bidirectional edge connecting node1i and node2i in the graph.
The distance between two nodes a and b is defined to be the minimum number of edges that are needed to go from a to b.
Return an integer array answer of size n, where answer[i] is the minimum distance between the ith node and any node in the cycle.
Example 1:
Input: n = 7, edges = [[1,2],[2,4],[4,3],[3,1],[0,1],[5,2],[6,5]]
Output: [1,0,0,0,0,1,2]
Explanation:
The nodes 1, 2, 3, and 4 form the cycle.
The distance from 0 to 1 is 1.
The distance from 1 to 1 is 0.
The distance from 2 to 2 is 0.
The distance from 3 to 3 is 0.
The distance from 4 to 4 is 0.
The distance from 5 to 2 is 1.
The distance from 6 to 2 is 2.
Example 2:
Input: n = 9, edges = [[0,1],[1,2],[0,2],[2,6],[6,7],[6,8],[0,3],[3,4],[3,5]]
Output: [0,0,0,1,2,2,1,2,2]
Explanation:
The nodes 0, 1, and 2 form the cycle.
The distance from 0 to 0 is 0.
The distance from 1 to 1 is 0.
The distance from 2 to 2 is 0.
The distance from 3 to 1 is 1.
The distance from 4 to 1 is 2.
The distance from 5 to 1 is 2.
The distance from 6 to 2 is 1.
The distance from 7 to 2 is 2.
The distance from 8 to 2 is 2.
Constraints:
3 <= n <= 105edges.length == nedges[i].length == 20 <= node1i, node2i <= n - 1node1i != node2i- The graph is connected.
- The graph has exactly one cycle.
- There is at most one edge between any pair of vertices.
Solutions#
Solution 1: Topological Sorting#
We can first convert the edges in $edges$ into an adjacency list $g$, where $g[i]$ represents all adjacent nodes of node $i$, represented as a set.
Next, we delete nodes layer by layer from the outside to the inside until only a cycle remains. The specific method is as follows:
We first find all nodes with a degree of $1$ and remove these nodes from the graph. If after deletion, the degree of its adjacent node becomes $1$, then we add it to the queue $q$. During this process, we record all deleted nodes in order as $seq$; and we use an array $f$ to record the adjacent node of each node that is closer to the cycle, i.e., $f[i]$ represents the adjacent node of node $i$ that is closer to the cycle.
Finally, we initialize an answer array $ans$ of length $n$, where $ans[i]$ represents the minimum distance from node $i$ to any node in the cycle, initially $ans[i] = 0$. Then, we start traversing from the end of $seq$. For each node $i$, we can get the value of $ans[i]$ from its adjacent node $f[i]$, i.e., $ans[i] = ans[f[i]] + 1$.
After the traversal, return the answer array $ans$.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes.
Similar problems:
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| class Solution:
def distanceToCycle(self, n: int, edges: List[List[int]]) -> List[int]:
g = defaultdict(set)
for a, b in edges:
g[a].add(b)
g[b].add(a)
q = deque(i for i in range(n) if len(g[i]) == 1)
f = [0] * n
seq = []
while q:
i = q.popleft()
seq.append(i)
for j in g[i]:
g[j].remove(i)
f[i] = j
if len(g[j]) == 1:
q.append(j)
g[i].clear()
ans = [0] * n
for i in seq[::-1]:
ans[i] = ans[f[i]] + 1
return ans
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| class Solution {
public int[] distanceToCycle(int n, int[][] edges) {
Set<Integer>[] g = new Set[n];
Arrays.setAll(g, k -> new HashSet<>());
for (var e : edges) {
int a = e[0], b = e[1];
g[a].add(b);
g[b].add(a);
}
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
if (g[i].size() == 1) {
q.offer(i);
}
}
int[] f = new int[n];
Deque<Integer> seq = new ArrayDeque<>();
while (!q.isEmpty()) {
int i = q.poll();
seq.push(i);
for (int j : g[i]) {
g[j].remove(i);
f[i] = j;
if (g[j].size() == 1) {
q.offer(j);
}
}
}
int[] ans = new int[n];
while (!seq.isEmpty()) {
int i = seq.pop();
ans[i] = ans[f[i]] + 1;
}
return ans;
}
}
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| class Solution {
public:
vector<int> distanceToCycle(int n, vector<vector<int>>& edges) {
unordered_set<int> g[n];
for (auto& e : edges) {
int a = e[0], b = e[1];
g[a].insert(b);
g[b].insert(a);
}
queue<int> q;
for (int i = 0; i < n; ++i) {
if (g[i].size() == 1) {
q.push(i);
}
}
int f[n];
int seq[n];
int k = 0;
while (!q.empty()) {
int i = q.front();
q.pop();
seq[k++] = i;
for (int j : g[i]) {
g[j].erase(i);
f[i] = j;
if (g[j].size() == 1) {
q.push(j);
}
}
g[i].clear();
}
vector<int> ans(n);
for (; k; --k) {
int i = seq[k - 1];
ans[i] = ans[f[i]] + 1;
}
return ans;
}
};
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| func distanceToCycle(n int, edges [][]int) []int {
g := make([]map[int]bool, n)
for i := range g {
g[i] = map[int]bool{}
}
for _, e := range edges {
a, b := e[0], e[1]
g[a][b] = true
g[b][a] = true
}
q := []int{}
for i := 0; i < n; i++ {
if len(g[i]) == 1 {
q = append(q, i)
}
}
f := make([]int, n)
seq := []int{}
for len(q) > 0 {
i := q[0]
q = q[1:]
seq = append(seq, i)
for j := range g[i] {
delete(g[j], i)
f[i] = j
if len(g[j]) == 1 {
q = append(q, j)
}
}
g[i] = map[int]bool{}
}
ans := make([]int, n)
for k := len(seq) - 1; k >= 0; k-- {
i := seq[k]
ans[i] = ans[f[i]] + 1
}
return ans
}
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| function distanceToCycle(n: number, edges: number[][]): number[] {
const g: Set<number>[] = new Array(n).fill(0).map(() => new Set<number>());
for (const [a, b] of edges) {
g[a].add(b);
g[b].add(a);
}
const q: number[] = [];
for (let i = 0; i < n; ++i) {
if (g[i].size === 1) {
q.push(i);
}
}
const f: number[] = Array(n).fill(0);
const seq: number[] = [];
while (q.length) {
const i = q.pop()!;
seq.push(i);
for (const j of g[i]) {
g[j].delete(i);
f[i] = j;
if (g[j].size === 1) {
q.push(j);
}
}
g[i].clear();
}
const ans: number[] = Array(n).fill(0);
while (seq.length) {
const i = seq.pop()!;
ans[i] = ans[f[i]] + 1;
}
return ans;
}
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